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Wednesday, August 20, 2008

12 Volt 30 Amp PSUUsing a single 7812 IC voltage regulator and multiple outboard pass transistors, this power supply can deliver output load currents of up to 30 amps. The design is shown below:

Notes:The input transformer is likely to be the most expensive part of the entire project. As an alternative, a couple of 12 Volt car batteries could be used. The input voltage to the regulator must be at least several volts higher than the output voltage (12V) so that the regulator can maintain its output. If a transformer is used, then the rectifier diodes must be capable of passing a very high peak forward current, typically 100amps or more. The 7812 IC will only pass 1 amp or less of the output current, the remainder being supplied by the outboard pass transistors. As the circuit is designed to handle loads of up to 30 amps, then six TIP2955 are wired in parallel to meet this demand. The dissipation in each power transistor is one sixth of the total load, but adequate heat sinking is still required. Maximum load current will generate maximum dissipation, so a very large heat sink is required. In considering a heat sink, it may be a good idea to look for either a fan or water cooled heat sink. In the event that the power transistors should fail, then the regulator would have to supply full load current and would fail with catastrophic results. A 1 amp fuse in the regulators output prevents a safeguard. The 400mohm load is for test purposes only and should not be included in the final circuit. A simulated performance is shown below:Calculations:This circuit is a fine example of Kirchoff's current and voltage laws. To summarise, the sum of the currents entering a junction, must equal the current leaving the junction, and the voltages around a loop must equal zero. For example, in the diagram above, the input voltage is 24 volts. 4 volts is dropped across R7 and 20 volts across the regulator input, 24 -4 -20 =0. At the output :- the total load current is 30 amps, the regulator supplies 0.866 A and the 6 transistors 4.855 Amp each , 30 = 6 * 4.855 + 0.866. Each power transistor contributes around 4.86 A to the load. The base current is about 138 mA per transistor. A DC current gain of 35 at a collector current of 6 amp is required. This is well within the limits of the TIP2955. Resistors R1 to R6 are included for stability and prevent current swamping as the manufacturing tolerances of dc current gain will be different for each transistor. Resistor R7 is 100 ohms and develops 4 Volts with maximun load. Power dissipation is hence (4^2)/200 or about 160 mW. I recommend using a 0.5 Watt resistor for R7. The input current to the regulator is fed via the emitter resistor and base emitter junctions of the power transistors. Once again using Kirchoff's current laws, the 871 mA regulator input current is derived from the base chain and the 40.3 mA flowing through the 100 Ohm resistor. 871.18 = 40.3 + 830. 88. The current from the regulator itself cannot be greater than the input current. As can be seen the regulator only draws about 5 mA and should run cold.

Regulated 12 Volt Supply

Notes:This circuit above uses a 13 volt zener diode, D2 which provides the voltage regulation. Aprroximately 0.7 Volts are dropped across the transistors b-e junction, leaving a higher current 12.3 Volt output supply. This circuit can supply loads of up to 500 mA. This circuit is also known as an amplified zener circuit.

Unregulated Power Supply

A basic full wave rectified power supply is shown below. The transformer is chosen according to the desired load. For example, if the load requires 12V at 1amp current, then a 12V, 1 amp rated transformer would do. However, when designing power supplies or most electronic circuits, you should always plan for a worst case scenario. With this in mind, for a load current of 1 amp a wise choice would be a transformer with a secondary current rating of 1.5 amp or even 2 amps. Allowing for a load of 50% higher than the needed value is a good rule of thumb. The primary winding is always matched to the value of the local electricity supply.

Notes:An approximate formula for determining the amount of ripple on an unregulated supply is:

Dual Regulated Power Supply

Notes:In this circuit, the 7815 regulatates the positive supply, and the 7915 regulates the negative supply. The transformer should have a primary rating of 240/220 volts for europe, or 120 volts for North America. The centre tapped secondary coil should be rated about 18 volts at 1 amp or higher, allowing for losses in the regulator. An application for this type of circuit would be for a small regulated bench power supply.

Increasing Regulator Current

Notes:Although the 78xx series of voltage regulators are available with different current outputs, you can boostthe available current output with this circuit. A power transistor is used to supply extra current to the loadthe regulator, maintaining a constant voltage. Currents up to 650mA will flow through the regulator, abovethis value and the power transistor will start to conduct, supplying the extra current to the load. This shouldbe on an adequate heat sink as it is likely to get rather hot. Suppose you use a 12v regulator, 7812. Theinput voltage should be a few volts higher to allow for voltage drops. Assume 20 volts. Lets also assumethat the load will draw 5amps. The power dissipation in the transistor will be Vce * Ic or (20-12)*8=40watt.It may keep you warm in the Winter, but you will need a large heatsink with good thermal dissipation.

If you want to increase the output current with a negative regulator, such as the 79xx series, then the circuitis similar, but an NPN type power transistor is used instead.

Logic PSU with Overvoltage Protection

Description:

A simple 5 Volt regulated PSU featuring overvoltage protection.notes:The 5 volt regulated power supply for TTL and 74LS series integrated circuits, has to be very precise and tolerant of voltage transients. These IC's are easily damaged by short voltage spikes. A fuse will blow when its current rating is exceeded, but requires several hundred milliseconds to respond. This circuit will react in a few microseconds, triggered when the output voltage exceeds the limit of the zener diode.This circuit uses the crowbar method, where a thyristor is employed and short circuits the supply, causing the fuse to blow. This will take place in a few microseconds or less, and so offers much greater protection than an ordinary fuse. If the output voltage exceed 5.6Volt, then the zener diode will conduct, switching on the thyristor (all in a few microseconds), the output voltage is therefore reduced to 0 volts and sensitive logic IC's will be saved. The fuse will still take a few hundred milliseconds to blow but this is not important now because the supply to the circuit is already at zero volts and no damage can be done. The dc input to the regulator needs to be a few volts higher than the regulator voltage. In the case of a 5v regulator, I would recommend a transformer with secondary voltage of 8-10volts ac. By choosing a different regulator and zener diode, you can build an over voltag trip at any value. I have a simulated transient graph of this over voltage protection circuit in the Design section.

Nicad Battery ChargerNotes:This simple charger uses a single transistor as a constant current source. The voltage acrossthe pair of 1N4148 diodes biases the base of the BD140 medium power transistor. The base-emitter voltage of the transistor and the forward voltage drop across the diodes are relatively stable. The charging current is approximately 15mA or 45mA with the switch closed. Thissuits most 1.5V and 9V rechargeable batteries.The transformer should have a secondary rating of 12V ac at 0.5amp, the primary should be220/240volts for Europe or 120volts ac for North America.

Variable Power SupplyNotes:Using the versatile L200 voltage regulator, this power supply has independent voltage and current limits. The mains transformer has a 12volt, 2 amp rated secondary, the primary winding should equal the electricity supply in your country, which is 240V here in the UK. The 10k control is adjusts voltage output from about 3 to 15 volts, and the 47 ohm control is the current limit. This is 10mA minimum and 2 amp maximum. Reaching the current limit will reduce the output voltage to zero. Voltage and current regulation equations can be found at this page.